Question

Let $f\left(x\right)$ be a twice differentiable function and $f^{\prime \prime }\left(0\right)=5$. Then $\underset{x\to 0}{lim}\frac{3f\left(x\right)-4f\left(3x\right)+f\left(9x\right)}{x^2}$ is equal to

• A. 30
• B. 120
• C. 40
• D. none of these

Answer 1

The correct option is B 120

Let, $L=\underset{x\to 0}{lim}\frac{3f\left(x\right)-4f\left(3x\right)+f\left(9x\right)}{x^2}$
Using L-hospital's rule,
$L=\underset{x\to 0}{lim}\frac{3f^{\prime }\left(x\right)-12f^{\prime }\left(3x\right)+9f^{\prime }\left(9x\right)}{2x}$
It is still in indeterminate form, so again using L-hospital's rule,
$L=\underset{x\to 0}{lim}\frac{3f^{\prime \prime }\left(x\right)-36f^{\prime \prime }\left(3x\right)+81f^{\prime \prime }\left(9x\right)}{2}=\frac{3f^{\prime \prime }\left(0\right)-36f^{\prime \prime }\left(0\right)+81f^{\prime \prime }\left(0\right)}{2}$
$=\frac{3\times 5-36\times 5+81\times 5}{2}=120$